Question 12

If $\vec{X}, \vec{Y}, \vec{Z}$ are coplanar, their scalar triple product $[\vec{X} \vec{Y} \vec{Z}] = 0$.

Since $\vec{X}, \vec{Y}, \vec{Z}$ are linear combinations of $\vec{x}, \vec{y}, \vec{z}$, we have $[\vec{X} \vec{Y} \vec{Z}] = \begin{vmatrix} \alpha & \beta & -1 \\ -1 & \alpha & \beta \\ \beta & -1 & \alpha \end{vmatrix} [\vec{x} \vec{y} \vec{z}]$.

Calculating $[\vec{x} \vec{y} \vec{z}] = 1(6-1) - 2(4-3) + 3(2-9) = 5 - 2 - 21 = -18 \\nnot= 0$.

So, the determinant must be zero: $\alpha(\alpha^2 + \beta) - \beta(-\alpha - \beta^2) - 1(1 - \alpha\beta) = 0$.

$\alpha^3 + \alpha\beta + \alpha\beta + \beta^3 - 1 + \alpha\beta = 0 \implies \alpha^3 + \beta^3 + (-1)^3 - 3\alpha\beta(-1) = 0$.

This factors as $(\alpha + \beta - 1)(\alpha^2 + \beta^2 + 1 - \alpha\beta + \alpha + \beta) = 0$.

Since $\alpha, \beta > 0$ and $\alpha \not= \beta$, the second term is strictly positive.

Thus, $\alpha + \beta - 1 = 0 \implies \alpha + \beta = 1$.

The value of $\alpha + \beta - 3 = 1 - 3 = -2$.